Minimum Depth of Binary Tree

地址：点击打开链接

其实就是树遍历的变种问题，可以用深度和层次遍历，我选择了深度遍历，注意对根节点的判断，理由是如果根节点如果左右子树只有一颗不为空，则这个子树即为最短路径的一部分

答案：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def minDepth(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        return self.getDepth(root, 0)
    def getDepth(self,root,depth):
        if root == None:
            return depth
        if not root.left and not root.right:
            depth += 1
            return depth
        if root.left and not root.right:
            depth += self.getDepth(root.left, 1)
        if root.right and not root.left:
            depth += self.getDepth(root.right, 1)
        if root.left and root.right:
            depth += min(self.getDepth(root.left, 1), self.getDepth(root.right, 1))
        return depth